Chapter 16 solutions_to_exercises(engineering circuit analysis 7th)

71
Engineering Circuit Analysis, 7 th Edition Chapter Sixteen Solutions 10 March 2006 1. We have a parallel RLC with R = 1 kΩ, C = 47 μF and L = 11 mH. (a) Q o = R(C/L) ½ = 65.37 (b) f o = ω o / 2π = (LC) / 2π = 221.3 Hz (c) The circuit is excited by a steady-state 1-mA sinusoidal source: 10 -3 0 o A jωL -j/ ωC The admittance Y(s) facing the source is Y(s) = 1/R + 1/sL + sC = C(s 2 + s/RC + 1/LC)/ s so Z(s) = (s/C) / (s 2 + s/RC + 1/LC) and Z(jω) = (1/C) (jω) / (1/LC – ω 2 + jω/RC). Since V = 10 -3 Z, we note that |V| > 0 as ω 0 and also as ω . PROPRIETARY MATERIAL. © 2007 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

Transcript of Chapter 16 solutions_to_exercises(engineering circuit analysis 7th)

Page 1: Chapter 16 solutions_to_exercises(engineering circuit analysis 7th)

Engineering Circuit Analysis, 7th Edition Chapter Sixteen Solutions 10 March 2006

1. We have a parallel RLC with R = 1 kΩ, C = 47 μF and L = 11 mH. (a) Qo = R(C/L)½ = 65.37

(b) fo = ωo/ 2π = (LC)-½ / 2π = 221.3 Hz (c) The circuit is excited by a steady-state 1-mA sinusoidal source:

10-3∠0o A jωL -j/ ωC

The admittance Y(s) facing the source is Y(s) = 1/R + 1/sL + sC = C(s2 + s/RC + 1/LC)/ s so Z(s) = (s/C) / (s2 + s/RC + 1/LC) and Z(jω) = (1/C) (jω) / (1/LC – ω2 + jω/RC).

Since V = 10-3 Z, we note that |V| > 0 as ω → 0 and also as ω → ∞.

PROPRIETARY MATERIAL. © 2007 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

Page 2: Chapter 16 solutions_to_exercises(engineering circuit analysis 7th)

Engineering Circuit Analysis, 7th Edition Chapter Sixteen Solutions 10 March 2006

2. (a) R = 1000 Ω and C = 1 μF. Qo = R(C/L)½ = 200 so L = C(R/ Qo)2 = 25 μH (b) L = 12 fH and C = 2.4 nF R = Qo (L/ C)½ = 447.2 mΩ (c) R = 121.7 kΩ and L = 100 pH C = (Qo / R)2 L = 270 aF

PROPRIETARY MATERIAL. © 2007 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

Page 3: Chapter 16 solutions_to_exercises(engineering circuit analysis 7th)

Engineering Circuit Analysis, 7th Edition Chapter Sixteen Solutions 10 March 2006

3. We take the approximate expression for Q of a varactor to be

Q ≈ ωCjRp/ (1 + ω2 Cj2 Rp Rs)

(a) Cj = 3.77 pF, Rp = 1.5 MΩ, Rs = 2.8 Ω (b) dQ/dω = [(1 + ω2 Cj

2 Rp Rs)(Cj Rp) - ωCjRp(2ωCj2 Rp Rs)]/ (1 + ω2 Cj

2 RpRs) Setting this equal to zero, we may subsequently write

CjRp (1 + ω2 Cj2 Rp Rs) - ωCjRp(2ωCj

2 Rp Rs) = 0 Or 1 – ω2 Cj

2 Rp Rs = 0. Thus, ωo = (Cj2 RpRs)–½ = 129.4 Mrad/s = 21.00 MHz

Qo = Q(ω = ωo) = 366.0

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Page 4: Chapter 16 solutions_to_exercises(engineering circuit analysis 7th)

Engineering Circuit Analysis, 7th Edition Chapter Sixteen Solutions 10 March 2006

4. Determine Q for (dropping onto a smooth concrete floor): (a) A ping pong ball: Dropped twice from 121.1 cm (arbitrarily chosen). Both times, it bounced to a height of 61.65 cm.

Q = 2πh1/ (h1 – h2) = 12.82 (b) A quarter (25 ¢). Dropped three times from 121.1 cm. Trial 1: bounced to 13.18 cm Trial 2: bounced to 32.70 cm Trial 3: bounced to 16.03 cm. Quite a bit of variation, depending on how it struck. Average bounce height = 20.64 cm, so

Qavg = 2πh1/ (h1 – h2) = 7.574 (c) Textbook. Dropped once from 121.1 cm. Didn’t bounce much at all- only 2.223 cm. Since the book bounced differently depending on angle of incidence, only one trial was performed.

Q = 2πh1/ (h1 – h2) = 6.4

All three items were dropped from the same height for comparison purposes. An interesting experiment would be to repeat the above, but from several different heights, preferrably ranging several orders of magnitude (e.g. 1 cm, 10 cm, 100 cm, 1000 cm).

PROPRIETARY MATERIAL. © 2007 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

Page 5: Chapter 16 solutions_to_exercises(engineering circuit analysis 7th)

Engineering Circuit Analysis, 7th Edition Chapter Sixteen Solutions 10 March 2006

5.

2 2

80Np/s, 1200 rad/s, Z( 2 ) 400

1200 80 1202.66 rad/s Q 7.5172

( )( ) ( )( 2 )Now, Y( ) C Y( 2 ) C2

80( 80 2400)Y( 160 1200) C Y( 160 1200)160 1200

d d

oo o

d d dd

d

j

s j s j js js j

jj jj

α ω α ω

ωωα

α ω α ω α α ωα ωα ω

= = − + = Ω

= + = ∴ = =

+ − + + − − += ∴ − + =

− +

− − +∴ − + = ∴ − + =

− +

2

1 180C400 2 15

1 229

30

C32,000 901

j

1 115.775 F; L 43.88mH; R 396.7C 2o

j

ω α−

− +=

− +

∴ = = Ω= = = =

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Page 6: Chapter 16 solutions_to_exercises(engineering circuit analysis 7th)

Engineering Circuit Analysis, 7th Edition Chapter Sixteen Solutions 10 March 2006

6.

2

2

2 6 2 2 2 6

3 5 3 2

1 1 2 0.1Y 0.2 0.22 0.1 1 1000 / 4 0.01 1000 10

2 0.1 1000 0.1 10000.2 04 0.1 10 4 0.01 100.1 10 4000 10 9.9 96,000 98.47 rad/s

−= + + = + +

+ + +

− + −= + + ∴ + =

+ + + +

+

∴ + = + ∴ = ∴

inj jω

j j jj j

=

ωω ω ω

ω ω ω ω ωω ω ω ω

ω ω ω ω ω ω

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Engineering Circuit Analysis, 7th Edition Chapter Sixteen Solutions 10 March 2006

7. 6 6Parallel: R 10 , L 1, C 10 , I 10 0 As μ−= = = = ∠ °

(a) 3 6 61 1000 rad/s; Q RC 10 1000LCo o oω ω + −= = = = =

(b) 6 6 5 3 3

2 2

26

1 I 100010 10 , V 10 /10 10Y 1000

10 10V , V1000 10000.001 101000 1000

j j

j

ωω ω

ω ωω ω

− − − − −

− −

⎡ ⎤⎛ ⎞ ⎛ ⎞− = = + −⎜ ⎟ ⎜ ⎟= +

ω V

995 996 997 998 999 1000 1001 1002 1003 1004 1005 999.5 1000.5

0.993 1.238 1.642 2.423 4.47 10.0 4.47 2.428 1.646 1.243 0.997 7.070 7.072

Y ⎢ ⎥⎝ ⎠ ⎝ ⎠⎣ ⎦

∴ = =⎛ ⎞ ⎛ ⎞+ − + −⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

PROPRIETARY MATERIAL. © 2007 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

Page 8: Chapter 16 solutions_to_exercises(engineering circuit analysis 7th)

Engineering Circuit Analysis, 7th Edition Chapter Sixteen Solutions 10 March 2006

8. (a)

2 2

42 6 2 2

2 2 6

5(100 / ) 0.1

6

5 (100 / ) 10 0.01500 10 100 10 100(20 ) 10 (1000 )2 2 2

100 5 1000 20 1000 400 10100 10 0 10 100 40,000, 99 960,000

400 10960,000 / 99

= + ++ +

Z 2

(b) 2

2 2 6

102000Z ( ) 2 2.294400 10

oin o

o o

ωωω ω

= + + =+ +

Ω

9

− −= + + = + + = + +

+ + + + + +

−∴ + = ∴ + = + =

+ +∴ = =

in

o

j jj j

j j j j jj j j j

ω ωω ω

ω ω ω ωω ω ω ω ω ω

ω ω ω ω ωω ωω 8.47 rad/s

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Page 9: Chapter 16 solutions_to_exercises(engineering circuit analysis 7th)

Engineering Circuit Analysis, 7th Edition Chapter Sixteen Solutions 10 March 2006

9. (a)

1 1 2 2 2

2

50 , 1000 1,002,500 1001.249

1 10LC 1,002,500

d o d o

o

s s ω α ω ω

ω α

− −

+

= = ∴ = + = ∴ =

= = = = = = Ω6 61 100.9975 H; R 10

2 C 100k

(b) 4 6 1 1Y 10 10 , 1000 Z 9997 1.43210.9975 Y

j ω ωω

− −⎛ ⎞= + − = ∴ = = ∠ °Ω ⎜ ⎟⎝ ⎠

α ω

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Page 10: Chapter 16 solutions_to_exercises(engineering circuit analysis 7th)

Engineering Circuit Analysis, 7th Edition Chapter Sixteen Solutions 10 March 2006

10.

min max

o

3 3

max min

214

max min max 3

3 3

535kHz, 1605kHz, Q 45 at one end andQ 45 for 535 1605 kHz

1 11/ 2 LC 535 10 ,1605 102 L C 2 L C

1L / L 3; L C 8.8498 102 535 10

RC 45,535 10 1605 10 . Use2

o

o

oo

f

f ff

ππ π

πωωπ

= = =≤ ≤ ≤

= ∴ × = × =

⎛ ⎞∴ = = = ×⎜ ⎟× ×⎝ ⎠

≤ × ≤ ≤ × max

3 3

14max

max min12

2 1605 10 20 10 C 45 C 223.1pFL8.8498 10L 397.6 H, L 44.08 H

223.1 10 9

π

μ μ−

∴ × × × × = ∴ =

×∴ = = = =

×

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Page 11: Chapter 16 solutions_to_exercises(engineering circuit analysis 7th)

Engineering Circuit Analysis, 7th Edition Chapter Sixteen Solutions 10 March 2006

11. (a) A 4

4 5 4 83

8 2 44 8

8 2 4

pply 1V. I 10 A1Y I 10 (1 [10 ( 10 )])10

4.4 101000 48.4 10 4.4 10 1000Y 10 11 104.4 4.4

1000 48.4 10 4.4 10Y ( )4.4

− − −−

− −− −

− −

± ∴ = − (b)

∴ = = + + − −×

× + × +∴ = + + × =

− × + ×=

R

in in

in

in

ss

s sss s

jjj

ω ωωω

8 2

14

At , 1000 48.4 10 , 45.45 krad/s

4.4 10Z ( ) 104.4

− −

−−

= = × =

⎛ ⎞×= = Ω⎜ ⎟

⎝ ⎠

o o o

oin o

o

jj kj

ω ω ω ω

ωωω

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Page 12: Chapter 16 solutions_to_exercises(engineering circuit analysis 7th)

Engineering Circuit Analysis, 7th Edition Chapter Sixteen Solutions 10 March 2006

12. 01 24 4.9 rad/sLC

ω = = = or f0 = 0

2ωπ

= 780 mHz

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Page 13: Chapter 16 solutions_to_exercises(engineering circuit analysis 7th)

Engineering Circuit Analysis, 7th Edition Chapter Sixteen Solutions 10 March 2006

13. ( )

06

1 1 200 rad/s1 25 10

1.01LC

ω−

= = =×

or f0 = 0

2ωπ

= 31.99 Hz

PROPRIETARY MATERIAL. © 2007 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

Page 14: Chapter 16 solutions_to_exercises(engineering circuit analysis 7th)

Engineering Circuit Analysis, 7th Edition Chapter Sixteen Solutions 10 March 2006

14. (a) 31 1 10 5

2 2 200R

RC Cα

α= ∴ = = = Ω

0 6

1 1 1000 rad/s10LC

ω−

= = = or f0 = 0

2ωπ

= 159.2 Hz

Zin(ω0) = R = 5 Ω (b) We see from the simulation result that the ratio of the test source voltage to its current is 5 Ω at the resonant frequency; the small error is due to the series resistance PSpice required.

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Page 15: Chapter 16 solutions_to_exercises(engineering circuit analysis 7th)

Engineering Circuit Analysis, 7th Edition Chapter Sixteen Solutions 10 March 2006

15. (a) -11 = 50 s 2RC

α = and 2 20 5000 rad/sdω ω α= − =

Zin(ω0) = R so find R.

( )2 2 20

1 1 40 Fd

CL L

μω ω α

= = =+

.( )2 2

1 250 2 2(50)

dLR

C

ω α

α

+= = = Ω

(b) The resonant frequency is 01 5000 rad/s or 795.8 HzfLC

= = .

We see from the simulation result that the ratio of the test source voltage to its current is 250 Ω at the resonant frequency; the small error is due to the series resistance PSpice required.

PROPRIETARY MATERIAL. © 2007 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

Page 16: Chapter 16 solutions_to_exercises(engineering circuit analysis 7th)

Engineering Circuit Analysis, 7th Edition Chapter Sixteen Solutions 10 March 2006

16. 1000 rad/s, Q 80, C 0.2 Fo oω μ= = =

(a) 6

2 6 3 6

1 10 80L 5H, Q RC R 400C 0.2 10 10 0.2 10o o

o

kωω −= = = = ∴ = = Ω

×× ×

(b) B = =

23

/ Q 1000 / 80 12.51 B 6.25 rad/s2

Z R / 1 400 10 / 1B / 2 6.25

o o

o oj

ω

ω ω ω ω

=

∴ =

− −⎛ ⎞∴ = + = × + ⎜ ⎟⎝ ⎠

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Page 17: Chapter 16 solutions_to_exercises(engineering circuit analysis 7th)

Engineering Circuit Analysis, 7th Edition Chapter Sixteen Solutions 10 March 2006

17.

1 2

21 2

o

32

1

103rad/s, 118,

Z( 105) 10

103 118

110.245110.245 , B 118 103 15 rad/s, Q 7.350B 15

7.350 1 17.350 RC RC 66.67 10 ,LC110.245 12,154

1 1Y( 105) 0.1 105C 15C 105CR 105L

o

oo

oo

j

j j j

ω ω

ω ω ω

ωω

ωω

++

−+

= =

= Ω

= = ×

∴ = = − = = = =

∴ = ∴ = = × = =

⎛ ⎞= = + − = +⎜ ⎟⎝ ⎠

12,154 C 18.456C105

0.1 1 1C 5.418mF, R C 12.304 , L 15.185 mH18.456 15 12,154C

⎛ ⎞− =⎜ ⎟⎝ ⎠

∴ = = = = Ω = =

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Page 18: Chapter 16 solutions_to_exercises(engineering circuit analysis 7th)

Engineering Circuit Analysis, 7th Edition Chapter Sixteen Solutions 10 March 2006

18. 30 krad/s, Q 10, R 600 ,= = =o o Ωω

(a) B 3 krad/sQ

o

o

ω = =

(b) 28 30N 1.3333B / 2 1.5

oω ω− −= = = −

(c) Zin(j28 000) = 600 / (1 – j1.333) = 360 ∠ 53.13o Ω (d)

1

1

Q1 1 10 (e)

Z ( 28,000) 28,000C ,C600 28,000L R 30,000 600

R 600 1 30,000 10 1 28 10 30 10L , ZQ 30,000 10 L 600 600 30 600 28 600

600Z28 301 1030 28

−⎡ ⎤= + − = =⎢ ⎥ ×⎣ ⎦

351.906 54.0903

−× ⎡ ⎤⎛ ⎞= = = ∴ = + × −⎜ ⎟⎢ ⎥× ⎝ ⎠⎣ ⎦

= = °Ω⎛ ⎞+ −⎜ ⎟⎝ ⎠

oin

o

ino o

in

j

j

j

ω

ω

j j

approx-true 360 351.906magnitude: 100% 100% 2.300%true 351.906

53.1301 54.0903angle: 100% 1.7752%54.0903

−= =

° − °= −

°

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Page 19: Chapter 16 solutions_to_exercises(engineering circuit analysis 7th)

Engineering Circuit Analysis, 7th Edition Chapter Sixteen Solutions 10 March 2006

19. 3400Hz, Q 8, R 500 , I 2 10 A B 50Hzo o Sf −= = = Ω = × ∴ = (a) 3 2 2 4002 10 500 / 1 N 0.5 1 N 4, N 3

50 / 2400 25 3 443.3 and 356.7 Hz

f

f

− −= ×V × + = ∴ + = = ± =

(b)

∴ = ± =

3 2 2

2

1 1I 0.5 10 1 N 4, N 15, N 15R 5001 N400 25 15

R

v

f

−= = × = × ∴ + = = = ±+

496.8 and 303.2 Hz= ± =∴

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Page 20: Chapter 16 solutions_to_exercises(engineering circuit analysis 7th)

Engineering Circuit Analysis, 7th Edition Chapter Sixteen Solutions 10 March 2006

20. 6 310 , Q 10, R 5 10 , . .= = = ×o o p rω

(a) 3

6

R 5 10Q L 0.L 10 1

5mHooω

×= ∴ =

×

0=

(b)

62

6

2

2 2 2 22 2

4 2 2

10 (c)

Approx: 2 5 / 1 N N 2.291 1.1146 Mrad/S10 / 20

1 1Exact: Y 1 Q 0.5 0.2 1 100 ( in Mrad/S)R

1 16.25 1 100( 2 1/ ), 2 0.0525, 2.0525

12.0525 1 0, 22

−= + ∴ = = ∴ =

⎡ ⎤⎛ ⎞ ⎛ ⎞= + − ∴ = + −⎢ ⎥⎜ ⎟ ⎜ ⎟⎝ ⎠⎝ ⎠⎣ ⎦

∴ = + − + − + = + =

− + = =

oo

o

j

ω ω

ωω ω ωω ω ω

ω ω ω ωω ω

ω ω ω ( )2.0525 2.0525 4 1.2569, 1.1211 Mrad/s+ − = =ω

1

22

1Approx: Y 30 tan N 30 , N 0.5774 , 1.0289 Mrad/s1/ 20

1 1 1Exact: Y 1 10 (in Mrad/s) tan 30 0.5774 105000

1 0.05774 0.05774 40.05774, 0.05774 1 0, 1.0293 Mrad/s2

− −∠ = ° ∴ = ° = = =

⎡ ⎤⎛ ⎞ ⎛ ⎞= + − ∴ ° = = −⎜ ⎟ ⎜ ⎟⎢ ⎥⎝ ⎠ ⎝ ⎠⎣ ⎦

+ +

ω

− = − − = = =

j

ω

ω ωω ω

ω ω ω ωω

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Page 21: Chapter 16 solutions_to_exercises(engineering circuit analysis 7th)

Engineering Circuit Analysis, 7th Edition Chapter Sixteen Solutions 10 March 2006

21.

(a) 6

4 8

1C 3 7 10nF 10 rad/s10 10

oω− −

= + = ∴ = =

(b) 6 8 3

9 6

33 3

1,0

9

R 10 10 5 5 10 50B / Q 20 krad/s

1 0Parallel current source is 3 10 At , I 10 3Z

V 3 10 5 10 15 90 V

o o

o o

o sj j

j

ωω

ω ω

− −

= = × == =

∠ °

Q C = × = ×

(c) 3

313

15 10 15 9015 10 N 1.5 V 8.321 33.69 V10 10 1 1.5o j

ω ω × ∠ °− = × ∴ = = ∴ = = ∠ °

× +

∴ = × × × = ∠ °

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Page 22: Chapter 16 solutions_to_exercises(engineering circuit analysis 7th)

Engineering Circuit Analysis, 7th Edition Chapter Sixteen Solutions 10 March 2006

22. (a)

6 6

6 2 6

2 4 6

2 6

6 2

6 2

6 2 6

(5 0.01 )(5 10 / ) (5 0.01 )(5 10 )Z ( )10 0.01 10 / 0.01 10 10

0.05 25 10 5 10Z ( )0.01 10 10

5 10 0.05 10,025Z ( )10 0.01 10

10,025 10At ,5 10 0.05 10 0.01

+ + + += =

+ + + ++ + + ×

=+ +

× − +

∴ =− +

= =× − −

in

in

in

o oo

o

s s s sss s s s

s s sss s

jjj

ω ωωω ω

ω ωω ωω

(b) 25 10,000Z ( ) (5 100) (5 100) 1002.510in oj j jω +

= + − = = Ω

9 2 7 22

2 9

, 10.025 10 100.25 5 10 0.5

99.75 9.975 10 , 10,000 rad/s

× − = × −

= × =

o oo

o o

ω ωω

ω ω∴

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Page 23: Chapter 16 solutions_to_exercises(engineering circuit analysis 7th)

Engineering Circuit Analysis, 7th Edition Chapter Sixteen Solutions 10 March 2006

23. , 1000Hz, Q 40, Z ( ) 2 B 25Hzo o in of j kω= = = Ω ∴ = 2000 1000, N , 1010, N 0.8

1 N 12.5−

= = ∴+

f fj

(a) Zin(jω) = Zin = 2000 / (1 + j0.8) = 1562 ∠ -38.66o Ω (b) 0.9 1.1 900 1100 Hzo of f f f< < ∴ < <

=

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Page 24: Chapter 16 solutions_to_exercises(engineering circuit analysis 7th)

Engineering Circuit Analysis, 7th Edition Chapter Sixteen Solutions 10 March 2006

24. Taking 2–½ = 0.7, we read from Fig. 16.48a: 1.7 kHz – 0.6 kHz = 1.1 kHz Fig. 16.48b: 2×107 Hz – 900 Hz = 20 MHz

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Page 25: Chapter 16 solutions_to_exercises(engineering circuit analysis 7th)

Engineering Circuit Analysis, 7th Edition Chapter Sixteen Solutions 10 March 2006

25. Bandwidth = 2 2 60 2 110f π ω ω= = − , where ω1 = ( ) 32 5.5 10π . π

(a) 2 1 Bω ω= + , therefore f2 = 5.5 + 103 kHz = 1.0055 MHz (b) ( )( )0 1 2 5.5 1005.5f f 74.37 kHz f= = =

(c) 3

00 6

74.37 1010

fQ = = 0.074 B

×=

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Page 26: Chapter 16 solutions_to_exercises(engineering circuit analysis 7th)

Engineering Circuit Analysis, 7th Edition Chapter Sixteen Solutions 10 March 2006

26. Bandwidth = 9210 Hz 1f f= − , where f1 = Hz. 675.3 10×

(a) 2 1f f B= + , therefore f2 = 1.0753 GHz

(b) ( )( )6 90 1 2 75.3 10 1.0753 10f= = × × =f f 284.6 MHz

(c) 6

00 9

284.6 1010

fQB

×= = = 0.2846

PROPRIETARY MATERIAL. © 2007 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

Page 27: Chapter 16 solutions_to_exercises(engineering circuit analysis 7th)

Engineering Circuit Analysis, 7th Edition Chapter Sixteen Solutions 10 March 2006

27. (a) To complete the sketch, we need to first find ω0, which we obtain in part (b). (b) 0 1 2 02000 rad/s or 318.3 Hzfω ω ω= = = (c) B = 2 1 3000 rad/s or 477.5 Hzω ω− =

(d) 0

2 1

2000 0.6673000

Q ωω ω

= = =−

PROPRIETARY MATERIAL. © 2007 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

Page 28: Chapter 16 solutions_to_exercises(engineering circuit analysis 7th)

Engineering Circuit Analysis, 7th Edition Chapter Sixteen Solutions 10 March 2006

28. (a) We begin by labelling the series string with the capacitor as string 1, and the other as string 2. We next find the parallel equivalent of each, and determine the frequency where Xp1 + Xp2 = 0.

Then 1

2 21 1

1p

R XXX+

= , and similarly 2

2 22 2

2p

R XXX+

= .

For 1 2

0 we have p pX X+ =2 2 2 2

1 1 2 2

1 2

0R X R XX X+ +

+ = [1]

At ω0, 10

1 XCω

= − ∴( )

242

222 201 1

121

0

105330

10330

R XX

ω

ω

++

=−

.

At ω0, 2 0 X Lω= ∴2 42 2

02 22

2 0

5 1010

R XX

2ωω

++= .

Enforcing Eq. [1], then, leads to ( )22 12

0 8 2

10 25 (330)10550.5 krad/s

(330)10 25(33)ω

−= =

or f0 = 87.61 kHz. (b) We see the simulation result agrees reasonably, with a resonant frequency of 87.6 kHz

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Page 29: Chapter 16 solutions_to_exercises(engineering circuit analysis 7th)

Engineering Circuit Analysis, 7th Edition Chapter Sixteen Solutions 10 March 2006

29. (a) We design for a bandwidth of 5.5 kHz, a low-frequency cut-off of 500 Hz, and a resonant impedance of 1 kΩ (no value was specified). Thus, we need to specify values for R, L, and C.

( )2 1

0 1 2

30

0 3

6 kHz

0.5 (6) 3 kHz

3 105.5 10

f f B

f f f

fQB

= + =

= = =

×= =

×

( )( )

00 0 3 3

0

1 so 28.9 nF5.5 10 2 10

QQ RC CR

ωω π

= = = =×

L = ( )

( )3 3

2 60

5.5 10 101 292 mH2 3 10Cω π

×= =

× and, of course, R = 1 kΩ

(b) From the simulation, we observe a bandwidth of 5.5 kHz, a lower frequency cutoff of approximately 500 Hz, and a peak impedance of 1000 Ω, as desired.

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Page 30: Chapter 16 solutions_to_exercises(engineering circuit analysis 7th)

Engineering Circuit Analysis, 7th Edition Chapter Sixteen Solutions 10 March 2006

30. (a) ( )( )0 6 6

1 1 1 1 4.38 kHz2 2 400 10 3.3 10

fLCπ π − −

= = =× ×

(b) 00

1 1 400 1.1010 3.3

L LQR RLC

ω= = = =

(c) Z at resonance = R = 10 Ω (d) Z at 0.438 kHz =

( )( ) ( )( )6

6

110 2 438 400 10 10 109.01 2 438 3.3 10

j jππ

−−

⎡ ⎤⎢ ⎥+ × − = −

×⎢ ⎥⎣ ⎦Ω

(e) Z at 43.8 kHz =

( )( ) ( )( )4

4

110 2 438 400 10 10 108.98 j j= −2 438 3.3 10

ππ

−−

⎡ ⎤⎢ ⎥+ × − Ω

×⎢ ⎥⎣ ⎦

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Page 31: Chapter 16 solutions_to_exercises(engineering circuit analysis 7th)

Engineering Circuit Analysis, 7th Edition Chapter Sixteen Solutions 10 March 2006

31. Bandwidth = 3 MHz, f1 = 17 kHz. (a) 2 1 3.017 MHzf f B= + = (b) 0 1 2 226.5 kHzf f f= =

(c) 00 0.0755fQ

B= =

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Page 32: Chapter 16 solutions_to_exercises(engineering circuit analysis 7th)

Engineering Circuit Analysis, 7th Edition Chapter Sixteen Solutions 10 March 2006

32. (a) Z0 = 1 Ω by definition

(b) 3

01 10 707 rad/s = 112.5 Hz

2LCω = = =

(c) PSpice simulation verifies an impedance of 1 Ω at f = 112.6 Hz.

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Page 33: Chapter 16 solutions_to_exercises(engineering circuit analysis 7th)

Engineering Circuit Analysis, 7th Edition Chapter Sixteen Solutions 10 March 2006

33. (a) Z0 = 1 kΩ by definition

(b) 6

01 10 707 krad/s = 112.5 kHz

2LCω = = =

(c) PSpice simulation verifies an impedance of 1 kΩ at f = 112.8 kHz.

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Page 34: Chapter 16 solutions_to_exercises(engineering circuit analysis 7th)

Engineering Circuit Analysis, 7th Edition Chapter Sixteen Solutions 10 March 2006

34. (a)

2

20A 6 , 3 6 2, 40V in series with 2 1 3

L1 6010 rad/s, Q 20R 3LC

10 1B 0.5, B 0.25, V ( ) 40Q 800V20 2

10V ( ) 800 / 10.25

oo o

out o o

out

j

j

ωω

ω

ωω

Ω = + = Ω

= = = = = Ω

= = = = =

−⎛ ⎞

∴ = + ⎜ ⎟

⎝ ⎠ (b)

9 rad/s800(Approx: V ( 9) 194.03V

1740 600Exact: V

3 (6 600 / )24,000V ( 9)

9[3 (54 66.67)]

out

out

out

j

j j

jj

204.86 13.325 V

ω

ω ω ω

= =

= ×+ −

=

∴ = = ∠ −+ −

PROPRIETARY MATERIAL. © 2007 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

Page 35: Chapter 16 solutions_to_exercises(engineering circuit analysis 7th)

Engineering Circuit Analysis, 7th Edition Chapter Sixteen Solutions 10 March 2006

35. 7Series: R 50 , L 4mH, C 10−= Ω = = (a) 3 71/ 4 10 50 krad/soω − −= × = (b) 350 10 / 2 7.958kHzof π= × =

(c) 3 3L 50 10 4 10Q 4

R 50o

oω −× × ×

= = =

(d) 3B / Q 50 10 / 4 12.5 krad/so oω= = × =

(e) 21 1 (1/ 2 ) 1/ 2 50 1 1/ 64 1/ 8 44.14 krad/so o oQ Qω ω ⎡ ⎤ ⎡ ⎤= + − = + − =⎣ ⎦⎣ ⎦

(f) 2 50 65 / 64 1/ 8 56.64 krad/sω ⎡ ⎤= + =⎣ ⎦

(g) 7 3Z ( 45,000) 50 (180 10 / 45) 50 42.22 65.44 40.18in j j j−= + − = − = ∠ − °Ω (h) 7

45,000Z / Z 10 / 45,000 50 4.444c R j= × =

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Page 36: Chapter 16 solutions_to_exercises(engineering circuit analysis 7th)

Engineering Circuit Analysis, 7th Edition Chapter Sixteen Solutions 10 March 2006

36. Apply 1 A, in at top. V 10VR∴ = (a)

8 83 3

3 8 3 8

2 11

10 1.2 10V Z 10 10 (0.5 10 1) 10 105

Z ( ) 10 (10 1.2 10 / ) 10 1.2 10 /

1.2 10 ,

in in

in o o

o o

s ss s

j j

(b) 3 3L 346.4 10Q 34.64

R 10o

oω −×

= = =

346.4 krad/s

ω ω ω ω ω

ω ω

− −

− −

×+ + × + = + +

= + − × ∴ = ×

= =

= × =∴

PROPRIETARY MATERIAL. © 2007 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

Page 37: Chapter 16 solutions_to_exercises(engineering circuit analysis 7th)

Engineering Circuit Analysis, 7th Edition Chapter Sixteen Solutions 10 March 2006

37. Find the Thévenin equivalent seen by the inductor-capacitor combination:

11 1 1

1

6

max

VSC : 1.5 V 10 0.105V V 50V125

50I 0.4A125

1.5OC :V 0 V 1.5V R 3.750.4

1000 41/ 4 0.25 10 1000,Q 1066.73.75

1000 1B / Q 0.9375, B 0.4688 rad/s1066.7 2

V Q V 1066.7 1.5 1600V

⎛ ⎞= + − ∴ =⎜ ⎟⎝ ⎠

∴↓ = =

= ∴ = ∴ = = Ω

×∴ = × × = = =

= = = =

= = × =

SC

OC th

o o

o o

C o th

ω

ω

Therefore, keep your hands off!

To generate a plot of |VC| vs. frequency, note that VC(jω) =

CjLj

Cj

ωω

ω

−+

75.35.1

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Page 38: Chapter 16 solutions_to_exercises(engineering circuit analysis 7th)

Engineering Circuit Analysis, 7th Edition Chapter Sixteen Solutions 10 March 2006

38. ,0Series, 500Hz, Q 10, X 500o o Lf = = = Ω

2 2

,0

1 2(a) (b)

500 = =L 2 (500)L L 0.15915 H, C 0.6366 FL (2 500)

X 500Q 10 R 50R R

+∴ = = = =×

= = = ∴ = Ω

oo

Lo

πω π μω π

6

6

1 10 0.5 250,0001 = +I 50 2 I 502 2

10 0.5I 1/ 50 ( 250,000 / ), V I2

250,000 /V V (2 450) 4.757 V50 ( 250,000 / )

V (2 500) 10,000 V V (2 550) 4.218V

⎛ ⎞ ⎛ ⎞×× − = + −⎜ ⎟ ⎜ ⎟

⎝ ⎠⎝ ⎠×

∴ = + − =

−= ∴ × =

+ −

× = × =

c

C c

c c

j f j j f jf f

j f fj f

j fj f f

πππ π

ππ

π

π π

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Page 39: Chapter 16 solutions_to_exercises(engineering circuit analysis 7th)

Engineering Circuit Analysis, 7th Edition Chapter Sixteen Solutions 10 March 2006

39.

1 4

8 2

8

1

X : 0, ,0 : 20,000 80,000 , Z ( 10 ) 20 0 SERIES120,000, 80,000 (64 4)10 82,462 rad/s, 68 10

LCR R 1 L 68 10 120,000 40,000, 170,000; Z( ) R L2L L LC R 40,000 C

8

20 R 10,

−= ∞ = − ± − = − + Ω ∴

= = ∴ = + = = = ×

×= = ∴ = × = = = + +

1000L10,00

∴− = −

in

d o o

s s j s j

α ω ω ω

α σ σ

σ

1 170,000R R R R 1.23080C 4 10,000

130.77 H, C 4.779 F170,000 1.2308

= − − ∴ = Ω

1.2308L40,000

∴ = = = =×

μ μ

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Page 40: Chapter 16 solutions_to_exercises(engineering circuit analysis 7th)

Engineering Circuit Analysis, 7th Edition Chapter Sixteen Solutions 10 March 2006

40. = 4.287 ∠ 59.04o kΩ

5 33 7 5

25 7

5 7

3

101/ 10 10 rad/s, Q 100, R 10,0001

1Q 500, R 500 0.2 50,00010 0.2

50 10 8.333 Q CR 10 8333 83.33100,000B 1200 rad/s, Z ( ) 8333

83.33(99 100)1099,000 N 1.6667, Z

600

−− −

ω = = = = Ω

= = = × = Ω×

= Ω ∴ = ω = × =

= = ω = Ω

−ω = ∴ = = −

o L PL

c PC

o o

in o

i

k

j

8.333( 99,000)1 1.667

=−n j

j

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Page 41: Chapter 16 solutions_to_exercises(engineering circuit analysis 7th)

Engineering Circuit Analysis, 7th Edition Chapter Sixteen Solutions 10 March 2006

41. Req = Qo/ ωo C = 50 / 105-7 = 5000 Ω. Thus, we may write 1/5000 = 1/8333 + 1/Rx so that Rx = 12.5 kΩ.

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Page 42: Chapter 16 solutions_to_exercises(engineering circuit analysis 7th)

Engineering Circuit Analysis, 7th Edition Chapter Sixteen Solutions 10 March 2006

42.

3 5

3 42

3 4

4 3

2

6

4

13mH 1.5mH 1mH, 2 F 8 F 10 F, 10 krad/s10

3 10 10Q 100, R 100 0.3 30.3

1.5 10 10Q 60, R 60 0.25 9000.25

692.3900 3000 692.3 Q 69.2310

692.3R 0.1444469.2310Q 125, R

10 0.1 8

− −

= μ + μ = μ ∴ω = =

× ×= = = × = Ω

× ×= = = × = Ω

= Ω ∴ = =

∴ = = Ω

= =× ×

o

p

p

L

LS

pc

k

2

4 52

, min

125 0.1 1562.5 10 F

1562.5Q 10 10 15625 156.25 R 0.064(156.25)

R 0.14444 0.064 0.2084 Z , 10 krad/s

= × = Ω μ

∴ = × × = ∴ = = Ω

∴ = + = Ω = ω =

c SC

S tot in o

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Page 43: Chapter 16 solutions_to_exercises(engineering circuit analysis 7th)

Engineering Circuit Analysis, 7th Edition Chapter Sixteen Solutions 10 March 2006

43. (a) 3

2

2

2

3

1/ 2 0.2 10 50 rad/s

Q 50 2.5 / 2 62.5, 2 62.5 7812.5

50 10Q 50, 10 50 2510

1000Q 100, 100 1 10 , R 7.8125 25 10 373150 0.2 1

50 1Q 50 3731 0.2 10 37.31; B 1.3400, B 0.670037.31 2

V 10

ω × × =

= × = × = Ω

×= = × = Ω

= = × = Ω = = Ω× ×

= × × × = = = =

(b)

∴ =

o

leftL

rightL

c p

o

o

k

k

3 3731− × = 3.731V

3

3

V = +10 [(2 125) (10 500) (1 100)]

101 1 1

2 125 10 500 1 100

+ −

= = ∠ − °+ +

+ + −

j j j

j j j

3.7321 0.3950 V+

3.731 V

2.638 V

50

↔ 1.34 rad/s

|V| (volts)

ω (rad/s)

PROPRIETARY MATERIAL. © 2007 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

Page 44: Chapter 16 solutions_to_exercises(engineering circuit analysis 7th)

Engineering Circuit Analysis, 7th Edition Chapter Sixteen Solutions 10 March 2006

44. (a) 6 3

42

,

,

1000 2000 rad/s, Q 2000 2 10 25 10 1000.25

R 20 10R 25,000 /100 2.5 ; Q 40L 2000 0.25

20,000R 12.5 R 12.5 2.5 151600

2000 0.25 1Q 33.33 V 1 33.33 16.667 V15 2

−ω = = × × × × =

×

∴ = = Ω = = =ω ×

∴ = = Ω ∴ = + = Ω

× (b)

∴ = = ∴ = ×

o c

C S Lo

L S tot

o x × =

20,000 50020,000 500 12,4922 499.68820,000 500

25,000( 250)25,000 250 2.4998 249.97525,000 250

Z 12.4922 2.4998 499.688 250 249.975 14.9920 0.2870

I 1/ 14.9920 0.2870 66.6902mA V 250

×= = + Ω

+−

− = = −−

∴ = + + − − = − Ω

= − = ∴ = ×in

x

jj jj

jj jjj j j j

j 366.6902 10−× = 16.6726V∴

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Page 45: Chapter 16 solutions_to_exercises(engineering circuit analysis 7th)

Engineering Circuit Analysis, 7th Edition Chapter Sixteen Solutions 10 March 2006

45. 2

2 2, , and 1 1

P PS S

R Q XQ CR R XQ Q

ω= = =+ +

2

2

1 1 1, S P S PS P

QX X C CC Cω ω

+= − = − ∴ =

Q

(a) ω = 103 rad/s, Q = 5 Therefore, RS = 5/26 = 192 Ω, CS = 26/25 μF = 1.06 μF (b) ω = 104 rad/s, Q = 50 Therefore, RS = 5/2501 = 2 Ω, CS = 2501/2500 μF = 1.0004 μF (c) ω = 105 rad/s, Q = 500 Therefore, RS = 5000/250001 = 20 mΩ, CS = 250001/250000 μF = 1.0 μF

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Page 46: Chapter 16 solutions_to_exercises(engineering circuit analysis 7th)

Engineering Circuit Analysis, 7th Edition Chapter Sixteen Solutions 10 March 2006

46. ( )2

22

1 1 , and P S P SQR R Q X X

Q+

= + =

2

21P SQC C

Q=

+

(a) ω = 103 rad/s, Q = 0.2 Therefore, RP = 5(1 + 0.04) = 5.2 kΩ, CP = 38.5 nF (b) ω = 104 rad/s, Q = 50 Therefore, RP = 5(1 + 0.0004) = 5.002 kΩ, CP = 400 pF (c) ω = 105 rad/s, Q = 500 Therefore, RP = 5(1 + 4×10–6) = 5 kΩ, CP = 4 pF

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Engineering Circuit Analysis, 7th Edition Chapter Sixteen Solutions 10 March 2006

47. 2 2

2 2, , and . 1 1

P PS S S

R Q XR QQ R X L LL Q Qω

= = = =+ + 21P Q+

(a) ω = 103 rad/s, Q = 142.4×103

Therefore, RS = 470/(1 + Q2) = 23.2 nΩ, LS = 3.3 μH (b) ω = 104 rad/s, Q = 14.24×103

Therefore, RS = 470/(1 + Q2) = 23.2 μΩ, LS = 3.3 μH (c) ω = 105 rad/s, Q = 1.424×103

Therefore, RS = 470/(1 + Q2) = 232 μΩ, LS = 3.3 μH

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Page 48: Chapter 16 solutions_to_exercises(engineering circuit analysis 7th)

Engineering Circuit Analysis, 7th Edition Chapter Sixteen Solutions 10 March 2006

48. ( )2

22

1 1 , and P S P SQR R Q X X

Q⎛ ⎞+

= + = ⎜ ⎟⎝ ⎠

2

2

1P S

QL LQ

⎛ ⎞+= ⎜ ⎟

⎝ ⎠

(a) ω = 103 rad/s, Q = 7.02×10–6

Therefore, RP = 470(1 + Q2) = 470 Ω, LP = 67 mF (b) ω = 104 rad/s, Q = 50 Therefore, RP = 470(1 + Q2) = 470 Ω, LP = 670 μF (c) ω = 105 rad/s, Q = 500 Therefore, RP = 470(1 + Q2) = 470 Ω, LP = 6.70 μF

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Page 49: Chapter 16 solutions_to_exercises(engineering circuit analysis 7th)

Engineering Circuit Analysis, 7th Edition Chapter Sixteen Solutions 10 March 2006

49. (a) For the left parallel circuit, 7 6

470 4710 10

RQLω −= ≈ = . Since Q > 5, the series

equivalent is a 10/47 Ω resistor in series with 1 μH. For the right parallel circuit, ( )7 810 10 200 20Q CRω −= ≈ = . Again, Q > 5, so the series equivalent is a 10/20 Ω = 500 mΩ resistor in series with 10 nF. We may therefore approximate the network as a 700 mΩ resistor in series with a 10 nF capacitor, in series with a 1 μH inductor, in series with the 10 μH inductor of interest. At the resonant frequency the network connected in series with the inductor has an impedance of 700 mΩ. The inductor present an impedance of 100 Ω. Thus, |Vx| = 1 V.

(b) ZL = ( ) 7 6470 ( 10 10 )

0.213 9.995 470 10

jj

j

= + Ω2

2

22

1

0.499 9.975 1L

Rj C j

Rj C

ω

ω+

. = = − Ω+

Z

Z3 = j100 Ω.

Thus, ( )3

1 3

1001 0 0.99745 0.0071 V0.714 0.02x

L

j jj

= ∠V = = ++ + +

ZZ Z Z

So that |Vx| = 0.99977 V . Our approximation was pretty accurate, at least at this frequency.

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Page 50: Chapter 16 solutions_to_exercises(engineering circuit analysis 7th)

Engineering Circuit Analysis, 7th Edition Chapter Sixteen Solutions 10 March 2006

50. 3

6

50 20 10(a) K = = 0.5 K 0.02100 10

1 0.59.82 H 0.5 9.82 24.55 H, 31.8 H 31.8 795 H0.02 0.02

2.572.57 nF0.5 0.02

×= =

∴ μ → × × = (b) same ordinate; divide numbers on abscissa by 50

257 nF→ =×

m f

μ μ → × = μ

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Engineering Circuit Analysis, 7th Edition Chapter Sixteen Solutions 10 March 2006

51. (a) 1 1Apply 1 V I 10A 0.5 I 5A ; 5A 0.2 can be replaced by 1 V in series with 0.2

1 ( 1) 2 4 20 20( 5)I 10 100.2 2 / 0.2 2 0.2 2 10

∴ = ∴ = ↓ Ω Ω

− − + + 10Z ( )20( 5)

+ ∴

(b) 2( / 5 10) 0.1( 50)K 2, K 5 Z ( )20( / 5 5) 25

+ += = ∴ → =

+ +m f ins sss s

(c) 1 10.1 0.2 , 0.2 0.4 , 0.5F 0.05F, 0.5I 0.5IΩ → Ω Ω → Ω → →

→= + = + = =+ + + +in

s s ss s s s

∴ =+ in

sss

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Engineering Circuit Analysis, 7th Edition Chapter Sixteen Solutions 10 March 2006

52. (a) 3 6 4

44 3 4

,8 , 2

4 3

2 2, ,4 6

4 6

1/ (2 8)10 10 10 rad/s

10Q 10 / 8 10 10 125 R 0.64125

10 10 102 8 10 mH Q 156.250.64

1R 0.64 156.25 15.625 ; Q 100, R 100 1 1010 10

R 20 15.625 10 4.673 Q 10 10 4

− −

(b)

(c) 6

6 1010 rad/s, Q stays the same, B 21.40 krad/s46.73

ω = ∴ = =o o

ω + =

= × = ∴ = = Ω

× ×+ = ∴ = =

∴ = × = Ω = = = × = Ω×

∴ = = Ω ∴ = × ×

o

L L S

L

L P C C P

P o

k k

k 3.673 10 46.73× =

6 4K = =f m10 /10 100, K 1 R s stay the same; 2 mH 20 H, 8mH 80 H,1 F 10nF′= ∴ → μ → μ μ →

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53.

3

0.1(a) K = =250, K 400 0.1F 1 F

250 4002 2505 1250 , 2H 1.25H, 4 I 10 I

400

∴ → = μ×

×Ω → Ω → = →

m f

x x (b) 3 6

1250

33

6 3 6 3

33 3

110 . Apply 1 V I 10 , I1250

1 101000 I 10 I1.25

1 0.8 0.8I 10 (1 10 ) 10 ; 1012500.8 10 1 1000I 10 0.2 10 Z 5 V

I 0.2

−−

− − −

−− −

ω =

0

∴ = ↓ =

∴ = ∴→ =

∴ = + + − = + =

×= + = × ∴ = = = − Ω =

x

x L

in

in th ocin

s

sss

s s s s js s

j j j kj j

1 μF

1.25 H

1250 Ω 103

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54. (a) I 2 0 A, 50 V 60 25 V= ∠ ° ω = ∴ = ∠ °s out (b) I 2 40 A, 50 V 60 65 V= ∠ ° ω = ∴ = ∠ °s out (c) I 2 40 A, 200, OTSK= ∠ ° ω = ∴s (d) K 30, I 2 40 A, 50 V 1800 65 V= = ∠ ° ω = ∴ = ∠ °m S out (e) K 30, K 4, I 2 40 A, 200 V 1800 65 V= = = ∠ ° ω = ∴ = ∠m f s out °

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55. (a) H /( ) 0.2 H 20 log 0.2 13.979dB= ∴ = = −dBs (b) H( ) 50 H 20 log 50 33.98dB= ∴ = =dBs (c) 12 26 6 13 292 380 (d) 37.6/ 20H 37.6dB H( ) 10 75.86= ∴ = =dB s (e) 8/ 20H 8dB H( ) 10 0.3981−= − ∴ = =dB s (f) 0.01/ 20H 0.01dB H( ) 10 1.0012= ∴ = =dB s

H j( 10) H 20 log 20 log 2 10 20 10 1 5 10 5 60 220

+= + ∴ = + =

+ + + + − +dBj

j j j j j6.451dB=

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56. (d) MATLAB verification- shown adjacent to Bode plots below.

(a) 20( 1) 0.2(1 )H( ) , 0.2 14dB100 1 /100+ +

= = → −+ +s ss

s s

1 10 100

(b) 2 2

2000( 1) 0.2 (1 )H( ) , 0.2 14dB( 100) (1 /100)

+ += = → −

+ +s s s ss

s s

(c) 2200 45 200 ( 5)( 40) 200(1 / 5)(1 / 40)( ) 45 , 200 46dB+ + + + + +

= + + = = = →s s s s s ss s

s s s sH

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57.

2

V (20 2 )(182 200 / ) 200 /H( )202 2 200 / 182 200 /

400( 10) 200(10 )2( 101 100) (1 )(100 )

20(1 /10)H( ) , 20 26dB(1 )(1 /100)

+ += = ×

+ + ++ +

= =+ + + +

+= →

+ +

C

R

s ss sI s s

s ss s s s

sss s

s

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58.

(a) 8

3 3

5 10 ( 100) 2.5 (1 /100)H( ) , 2.5 8dB( 20)( 1000) (1 / 20)(1 /1000)

× + += =

+ + + +s s s ss

s s s s→

(b) (c)

2 9

3 3

Corners: 20, 34dB;

100, 34dB;

1000, 54dB

Intercepts: 0dB, 2.5 1, 0.42.5 ( /100) 2.5 (20)101, 8dB; 0dB, 1 22,360 rad/s

( / 20)( /1000) 100

ω =

ω =

ω =

ω = ω =

ω ω ωω = = = ∴ω =

ω ω ωω

2

2 2 3

orners: 20, 31.13dB

1 ( /100)100, 36.69dB H 20 log 2.5

[1 ( / 20) ][1 ( /1000) ]1000, 44.99dB

ω =

+ ωω = = ω

+ ω + ω

ω =

dB

C

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59.

(a) 8

3 3

5 10 ( 100) 2.5 (1 /100)H( ) ,( 20)( 1000) (1 / 20)(1 /1000)

× + += =

+ + + +s s s ss

s s s s

(b) (c)

2 : 901010 : 90 45 45 log 58.520100 100100 : 90 45 45 log 45 45 log 58.520 100

200 200200 : 90 90 45 45 log 3 45 45 log 17.9100 100

1000 : 90 9

ω

ω

ω

ω

ω

= °

⎛ ⎞= ∠ = ° − ° + ° = °⎜ ⎟⎝ ⎠

⎛ ⎞ ⎛ ⎞= ∠ = ° − ° + ° + ° + ° = °⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

⎛ ⎞ ⎛ ⎞= ∠ = ° − ° + ° + ° − ° + ° = °⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

= ∠ = ° −

= ∠

10000 90 3 45 45 log 451000

10,000 : 90 90 90 3 90 180ω

⎛ ⎞° + ° − ° + ° = − °⎜ ⎟⎝ ⎠

= ∠ = ° − ° + ° − × ° = − °

1 1 1

1 1 1

1 1 1

1 1 1

1

2 : 90 tan 0.02 tan 0.1 3tan 0.002 85.09

10 : 90 tan 0.1 tan 0.5 3tan 0.01 67.43

100 : 90 tan 1 tan 5 3tan 0.1 39.18

200 : 90 tan 2 tan 10 3tan 0.2 35.22

1000 : 90 tan 10 t

ω = ∠

ω

ω

ω

ω

− − −

− − −

− − −

− − −

= ° + − − = °

= ∠ = ° + − − = °

= ∠ = ° + − − = °

= ∠ = ° + − − = °

= ∠ = ° + − 1 1

1 1 1

an 50 3tan 1 49.56

10,000 : 90 tan 100 tan 500 3tan 10 163.33ω

− −

− − −

− = − °

= ∠ = ° + − − = − °

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60. 2

2 2

2

2

20 400 20 400(a) (b) (c)

H( ) 1

1 2 0.5( / 20) ( / 20)400

20, 0.520 log 400 52dBCorrection at is 20 log 2 0 dB

o

o

s sss s s

s ss

ω ζ

ω ζ

+ += + + =

+ × +=

∴ = ==

=

5 : H 52 2 20 log 5 24.0dB(plot)

H 20 log 1 16 4 23.8dB (exact)

100 : H 0dB (plot)

H 20 log 1 0.04 0.2 0.170 dB (exact)

dB

dB

dB

dB

j

j

ω

ω

− × =

= − + =

= =

= − + = −

= =

Hdb

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61. (a)

225

V 25 25 0.025

(c) 0.520, H( 20) H 15.68 dB H( 20) 80.541 4 0.5 dB

jj jj

ω = = ∴ = − ∠ = −− +

°

H( )V 10 25 1000 / 10 25 1000 11 2

8 10 10110, 1/ 8 correction 20 log 2 12 dB8

0.025 32 dB

= = = =+ + + + ⎛ ⎞⎛ ⎞ ⎛ ⎞+ +⎜ ⎟⎜ ⎟ ⎜ ⎟

⎝ ⎠⎝ ⎠ ⎝ ⎠⎛ ⎞∴ = = ∴ = − × =⎜ ⎟⎝ ⎠

→ −

R

o

s sss s s s s s

ω ζ(b)

HdB ang(H)

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62. 3 6

1 2 3 3 6

1/(50 10 10 ) 201st two stages, H ( ) H ( ) 10; H ( )1/(200 10 10 ) 5

20 400H( ) ( 10)( 10)5 1 / 5

400 52 dB

s s ss s

ss s

− × × −= = − = =

+ × × +

− −⎛ ⎞∴ = − − =⎜ ⎟+ +⎝ ⎠− →

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63. (a) 20log10(0.1) = -20 dB (b) (c)

51 1 1

15 51

5 6

5 6

t stage: C 1 F, R , R 10 H (S) R C 0.11/ R C

2nd stage: R 10 , R 10 , C 1 F H ( )1/ R C

1/(10 10 ) 10H ( )1/(10 10 ) 10

3rd stage: same as 2nd1H( ) ( 0.1 )

A A fA A fA A

B fBB fB fB B

fB fB

B

s s

ss

ss s

s s

μ

μ

= = ∞ = ∴ = − = −

−= = = ∴ =

+

×

1s

∴ = = −+ × +

−∴ = − 2

0 10 0.110 10 (1 /10)

ss s s

−⎛ ⎞⎛ ⎞ = −⎜ ⎟⎜ ⎟+ + +⎝ ⎠⎝ ⎠

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64. An amplifier that rejects high-frequency signals is required. There is some ambiguity in the requirements, as social conversations may include frequencies up to 50 kHz, and echolocation sounds, which we are asked to filter out, may begin below this value. Without further information, we decide to set the filter cutoff frequency at 50 kHz to ensure we do not lose information. However, we note that this decision is not necessarily the only correct one.

Our input source is a microphone modeled as a sinusoidal voltage source having a peak amplitude of 15 mV in series with a 1-Ω resistor. Our output device is an earphone modeled as a 1-kΩ resistor. A voltage of 15 mV from the microphone should correspond to about 1 V at the earphone according to the specifications, requiring a gain of 1000/15 = 66.7.

If we select a non-inverting op amp topology, we then need 65.7 1- 66.7 1

==RR f

Arbitrarily choosing R1 = 1 kΩ, we then need Rf = 65.7 kΩ. This completes the amplification part. Next, we need to filter out frequencies greater than 50 kHz.

Placing a capacitor across the microphone terminals will “short out” high frequencies.

We design for ωc = 2πfc = 2π(50×103) = filtermicCR

1 . Since Rmic = 1 Ω, we require

Cfilter = 3.183 μF.

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65. We choose a simple series RLC circuit. It was shown in the text that the “gain” of the

circuit with the output taken across the resistor is ( )[ ] 2

122222 -1

RC CRLC

AV

ωω

ω

+= .

This results in a bandpass filter with corner frequencies at

LCLCCR-RC

Lc 24

22 ++=ω and

LCLCCRRC

Hc 24

22 ++=ω

If we take our output across the inductor-capacitor combination instead, we obtain the opposite curve- i.e. a bandstop filter with the same cutoff frequencies. Thus, we want

2π(20) = LC

LCCR-RC2

4 22 ++ and 2π(20×103) =

LCLCCRRC

24

22 ++

Noting that

Hcω – Lcω = R/L = 125.5 krad/s, we arbitrarily select R = 1 kΩ, so that L =

7.966 mH. Returning to either cutoff frequency expression, we then find C = 7.950 μF

PSpice verification. The circuit performs as required, with a lower corner frequency of about 20 Hz and an upper corner frequency of about 20 kHz.

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66. We choose a simple RC filter topology:

Where RC1

1 VV

in

out

ωj+= and hence

( )2in

out

RC1

1 VV

ω+= . We desire a cutoff

frequency of 1 kHz, and note that this circuit does indeed act as a low-pass filter (higher frequency signals lead to the capacitor appearing more and more as a short circuit). Thus,

( ) 21

RC1

1 2

c

=+

where ωc = 2πfc = 2000π rad/s.

A small amount of algebra yields 1 + [2π(1000)RC]2 = 2 or 2000πRC = 1. Arbitrarily setting R = 1 kΩ, we then find that C = 159.2 nF. The operation of the filter is verified in the PSpice simulation below:

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67. We are not provided with the actual spectral shape of the noise signal, although the reduction to 1% of its peak value (a drop of 40 dB) by 1 kHz is useful to know. If we place a simple high-pass RC filter at the input of an op amp stage, designing for a pole at 2.5 kHz should ensure an essentially flat response above 25 kHz, and a 3 dB reduction at 2.5 kHz. If greater tolerance is required, the 40 dB reduction at 1 kHz allows the pole to be moved to a frequency even closer to 1 kHz. The PSpice simulation below shows a

filter with R = 1 kΩ (arbitrarily chosen) and C = nF 63.66 )1000)(105.2(2

13 =

×π.

At a frequency of 25 kHz, the filter shows minimal gain reduction, but at 1 kHz any signal is reduced by more than 8 dB.

We therefore design a simple non-inverting op amp circuit such as the one below, which

with Rf = 100 kΩ and R1 = 1 kΩ, has a gain of 100 V/V. In simulating the circuit, a gain of approximately 40 dB at 25 kHz was noted, although the gain dropped at higher frequencies, reaching 37 dB around 80 kHz. Thus, to completely assess the suitability of design, more information regarding the frequency spectrum of the “failure” signals would be required.

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68. We select a simple series RLC circuit with the output taken across the resistor to serve as a bandpass filter with 500 Hz and 5000 Hz cutoff frequencies. From Example 16.12, we know that

(500)2 4LC CR2LC

12LR- 22 πω =++=

Lc

and

(5000)2 4LC CR2LC

12LR 22

Hπω =++=c

With -

LH cc ωω = 2p(5000 – 500) = R/L, we (arbitrarily) select R = 1 kΩ, so that L = 35.37 mH. Substituting these two values into the equation for the high-frequency cutoff, we find that C = 286.3 nF. We complete the design by selecting R1 = 1 kΩ and Rf = 1 kΩ for a gain of 2 (no value of gain was specified). As seen in the PSpice simulation results shown below, the circuit performs as specified at maximum gain (6 dB or 2 V/V), with cutoff frequencies of approximately 500 and 5000 KHz and a peak gain of 6 dB.

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69. For this circuit, we simply need to connect a low-pass filter to the input of a non- inverting op amp having Rf/R1 = 9 (for a gain of 10). If we use a simple RC filter, the cutoff frequency is

(3000)2 RC1 πω ==c

Selecting (arbitrarily) R = 1 kΩ, we find C = 53.05 nF. The PSpice simulation below shows that our design does indeed have a bandwidth of 3 kHz and a peak gain of 10 V/V (20 dB).

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70. We require four filter stages, and choose to implement the circuit using op amps to isolate each filter sub- circuit. Selecting a bandwidth of 1 rad/s (no specification was given) and a simple RLC filter as suggested in the problem statement, a resistance value of 1 Ω leads to an inductor value of 1 H (bandwidth for this type of filter = ωH – ωL = R/L). The capacitance is found by designing each filter’s respective resonant frequency ( LC1 ) at the desired “notch” frequency. Thus, we require CF1 = 10.13 μF, CF2 = 2.533 μF, CF3 = 1.126 μF and CF4 = 633.3 nF. The Student Version of PSpice® will not permit more than 64 nodes, so that the total solution must be simulated in two parts. The half with the filters for notching out 50 and 100 Hz components is shown below; an additional two op amp stages are required to complete the design.

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71. Using the series RLC circuit suggested, we decide to design for a bandwidth of 1 rad/ s (as no specification was provided). With ωH – ωL = R/ L, we arbitrarily select R = 1 Ω so that L = 1 H. The capacitance required is obtained by setting the resonant frequency of the circuit ( LC1 ) equal to 60 Hz (120π rad/s). This yields C = 7.04 μF.

1 H

1 Ω

7.04 μF

vin vout

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